Question

find the analytic function f(z)=u(x,y)+iv(x,y) such that u+v=(x-y)(x^2+4xy+y^2)

Answer 1

so this has been up a while without help. I'll give it a walk if you want. I think it is missing information though

Answer 2

wack*

Answer 3

not missing anything

Answer 4

If \(f\) is analytic, you know the Cauchy-Riemann equations are satisfied...

Answer 5

if we show it is harmonic, we can also use the laplace equations

Answer 6

\[\begin{align*}
u+v&=(x-y)(x^2+4xy+y^2)\\[2ex]\hline
\frac{\partial }{\partial x}\left[u+v\right]&=\frac{\partial}{\partial x}[\cdots]\\[1ex]
\frac{\partial u}{\partial x}+\frac{\partial v}{\partial x}&=x^2+4xy+y^2+(x-y)(2x+4y)\\[1ex]
&=3x^2+6xy-3y^2&(1)\\[2ex]\hline
\frac{\partial }{\partial y}\left[u+v\right]&=\frac{\partial}{\partial y}[\cdots]\\[1ex]
\frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}&=-(x^2+4xy+y^2)+(x-y)(4x+2y)\\[1ex]
&=3x^2-6xy-3y^2&(2)
\end{align*}\]
Since \(\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}\) and \(\dfrac{\partial v}{\partial x}=-\dfrac{\partial u}{\partial y}\), you get
\[\begin{align*}
2\frac{\partial u}{\partial x}+0\frac{\partial v}{\partial x}&=6x^2-6y^2&(1)+(2)\\[1ex]
\frac{\partial u}{\partial x}&=3x^2-3y^2\\[2ex]
0\frac{\partial u}{\partial x}+2\frac{\partial v}{\partial x}&=12xy\\[1ex]
\frac{\partial v}{\partial x}&=6xy\end{align*}\]

Answer 7

Whoops, those last two lines correspond to \((1)-(2)\).

Answer 8

By the C-R equations, you also get
\[\begin{cases}\dfrac{\partial u}{\partial x}=3x^2-3y^2&(1)\\[1ex]
\dfrac{\partial v}{\partial x}=6xy&(2)\end{cases}
\implies
\begin{cases}\dfrac{\partial u}{\partial y}=-6xy&(3)\\[1ex]
\dfrac{\partial v}{\partial y}=3x^2-3y^2&(4)\end{cases}\]
So, integrating \((1)\) with respect to \(x\), you get
\[\begin{align*}\int\frac{\partial u}{\partial x}\,dx&=\int(3x^2-3y^2)\,dx\\[1ex]
u&=x^3-3xy^2+g(y)\end{align*}\]
Differentiating with respect to \(y\), you get
\[\begin{align*}\frac{\partial u}{\partial y}&=-6xy+\frac{dg(y)}{dy}\\[1ex]
-6xy&=-6xy+\frac{dg}{dy}&\text{by }(3)\\[1ex]
\frac{dg}{dy}&=0\end{align*}\]
Solve for \(g\), and you can determine \(u(x,y)\). The same procedure is used for finding \(v(x,y)\).

Answer 9

thanks very much solved my problem